31958x^2+(362x^2)+(13248x^2)-(37x^2)=45

Solution for 31958x^2+(362x^2)+(13248x^2)-(37x^2)=45 equation:

31958x^2+(362x^2)+(13248x^2)-(37x^2)=45

We move all terms to the left:

31958x^2+(362x^2)+(13248x^2)-(37x^2)-(45)=0

We add all the numbers together, and all the variables

45531x^2-45=0

a = 45531; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·45531·(-45)
Δ = 8195580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8195580}=\sqrt{324*25295}=\sqrt{324}*\sqrt{25295}=18\sqrt{25295}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{25295}}{2*45531}=\frac{0-18\sqrt{25295}}{91062}
=-\frac{18\sqrt{25295}}{91062}
=-\frac{\sqrt{25295}}{5059}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{25295}}{2*45531}=\frac{0+18\sqrt{25295}}{91062}
=\frac{18\sqrt{25295}}{91062}
=\frac{\sqrt{25295}}{5059}
$